All Calculus 1 Resources
Example Questions
Example Question #1 : Lines
Find the equation of the line tangent to at .
To find the equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form , where m is the slope and is the y adjustment. To get the slope, find the derivative of and plug in the desired point for , giving us an answer of for the slope.
To find the y adjustment pick a point 0 in the original function. For simplicity, let's plug in , which gives us a y of 1, so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is
Now you simply solve for , which is .
Final equation of the line tangent to at is
Example Question #3 : Lines
Find the equation of the line tangent to at .
To get the slope, find the derivative of and plug in the desired point for , giving us an answer of for the slope.
Remember that the derivative of .
To find the adjustment pick a point (for example) in the original function. For simplicity, let's plug in , which gives us a of , so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is
The coefficient in front of the is the slope.
Now you simply solve for , which is .
Final equation of the line tangent to at is .
Example Question #4 : Lines
Find the equation of the line tangent to at .
The equation of the tangent line is To find , the slope, calculate the derivative and plug in the desired point.
The next step is to choose a coordinate on the original function. We can choose any value and calculate its value.
Let's choose .
The value at this point is .
Plugging in those values we can solve for .
Solving for we get =.
Example Question #3 : Lines
A function, , is given by
.
Find the line tangent to at .
First we need to find the slope of at . To do this we need the derivative of . To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.
At,
Now we need to know
.
Now we have a slope, and a point
so we can use the point-slope formula to find the equation of the line.
Plugging in and rearranging we find
.
Example Question #2 : Lines
Let .
Find the equation for a line tangent to when .
First, evaluate when .
Thus, we need a line that contains the point
Next, find the derivative of and evaluate it at .
To find the derivative we will use the power rule,
.
This indicates that we need a line with a slope of 8.
In point-slope form, , a line with the point and a slope of 8 will be:
Example Question #4 : Lines
What is the equation of the line tangent to at ? Round to the nearest hundreth.
The tangent line to at must have the same slope as .
Applying the chain rule we get
.
Therefore the slope of the line is,
.
In addition, the tangent line touches the graph of at . Since , the point lies on the line.
Plugging in the slope and point we get .
Example Question #2711 : Calculus
Find the equation of the tangent line, where
, at .
In order to find the equation of the tangent line at , we first find the slope.
To do this we need to find .
Since we have found , now we simply plug in 1.
Now we need to plug in 1, into , to find a point that the tangent line touches.
Now we can use point-slope form to figure out what the equation of the tangent line is at .
Remember that point-slope for is
where and is the point where the tangent line touches , and is the slope of the tangent line.
In our case, , , and .
Thus our tangent line equation at is
.
Example Question #1 : Equation Of Line
Find the equation of the tangent line of
, at .
In order to find the equation of the tangent line at , we first find the slope.
To do this we need to find using the power rule .
Since we have found , now we simply plug in 1.
Now we need to plug in 1, into , to find a point that the tangent line touches.
Now we can use point-slope form to figure out what the equation of the tangent line is at .
Remember that point-slope for is
where and is the point where the tangent line touches , and is the slope of the tangent line.
In our case, , , and .
Thus our tangent line equation at is
.
Example Question #11 : How To Find Equation Of Line By Graphing Functions
Give the general equation for the line tangent to at the point .
The equation of the tangent line has the form .
The slope can be determined by evaluating the derivative of the function at .
Plugging this into the point slope equation, we get
can be determined by evaluating the original function at .
Plugging this into the previous equation and simplifying gives us
Example Question #2714 : Calculus
Find the slope of the line tangent to the following function at .
None of these
To find the slope of the line tangent you must take the derivative of the function. The derivative of cosine is negative sine and the derivative of sine is cosine.
This makes the derivative of the function
.
Plug in the given x to get the slope.